I hope this is vaguely helpful. I put charges in brackets- sorry that it’s hard to read For i) initially you have V(2+) —->V(3+) This is feasible as E is >0 (1.52-(-0.26)=+1.78V)) Then, because you have excess MnO4- and the half cell potential of the MnO4- is still greater than for the vanadium (which is now all V(3+), V(3+) —> VO(2+) And in a similar way, VO(2+) —> VO2(+) Now all the vanadium is VO2+ and there is no reaction involving vanadium with a Eo greater than 0, so it will stay like that. The half equation is just the overall change in the vanadium- you start with V(2+) and finish with VO2(+) and balance it like any other half equation (lmk if you want me to explain)
For ii) basically I think the way my teacher said (can’t actually remember but it seems to work, even if the chemistry is wrong). If you add more Fe3+, more Fe3+ will be converted to Fe2+, so more electrons will be used up in the positive electrode, so there will be a greater p.d because the negative electrode still has the same number of electrons as before. I know that that is not very well explained. If there’s anything you think I can explain better please say